Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
string NumberToString(T Number) {
  std::ostringstream second;
  second << Number;
  return second.str();
}
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p % 1000000007LL;
    p = p * p % 1000000007LL;
    q >>= 1;
  }
  return f;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long const N = 2e5 + 53;
long long l[N], r[N], cum[N];
int n;
inline bool between(double x, double a, double b) { return (x >= a && x <= b); }
int tahtha(int e) {
  if (e < l[0]) return -1;
  int le = 0, ri = n - 1, mid;
  while (le < ri) {
    mid = (le + ri + 1) / 2;
    if (l[mid] <= e) {
      le = mid;
    } else {
      ri = mid - 1;
    }
  }
  return le;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(7);
  double Sy, a, b;
  cin >> Sy >> a >> b;
  cin >> n;
  cin >> l[0] >> r[0];
  cum[0] = r[0] - l[0];
  for (int i = 1; i < n; i++) {
    cin >> l[i] >> r[i];
    cum[i] = r[i] - l[i] + cum[i - 1];
  }
  int q;
  cin >> q;
  double x, y;
  for (auto i = 0; i < (q); ++i) {
    cin >> x >> y;
    double b1 = (a == x) ? a : (y * a - Sy * x) / (y - Sy);
    double b2 = (b == x) ? b : (y * b - Sy * x) / (y - Sy);
    auto istart = tahtha((int)b1);
    auto it_end = upper_bound(r + max(0, istart), r + n, (int)ceil(b2) - 1);
    int iend = (it_end == end(r)) ? -1 : (it_end - begin(r));
    double shadow = 0.0;
    if (istart >= 0 && between(b1, l[istart], r[istart]))
      shadow += min((double)r[istart], b2) - b1;
    if (iend >= 0 && istart != iend && between(b2, l[iend], r[iend]))
      shadow += b2 - l[iend];
    if (iend > istart) shadow += cum[iend - 1] - cum[istart];
    cout << (b - a) * (shadow) / abs(b2 - b1) << '\n';
  }
  return 0;
}